Answer
$$-9.50 \mathrm{MeV}$$
Work Step by Step
$$Q_{3}=\left(m_{\mathrm{as}_{\mathrm{v}}}-m_{\mathrm{s}_{\mathrm{m}} \mathrm{m}_{\mathrm{h}}}-m_{\mathrm{r}_{\mathrm{Hc}}}\right) c^{2}$$ $$=(235.0439 \mathrm{u}-232.0381 \mathrm{u}-3.0160 \mathrm{u})(931.5 \mathrm{MeV} / \mathrm{u}) $$ $$=-9.50 \mathrm{MeV}$$
